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    really confused why the grade boundary for june 2014 was so high, was one of the hardest papers
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    Ikr :'(
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    (Original post by Daniel9998)
    really confused why the grade boundary for june 2014 was so high, was one of the hardest papers
    What was it? I haven't checked yet :/

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    (Original post by lyricalvibe)
    What was it? I haven't checked yet :/

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    it was 70 for an A, thats not too bad considering the questions, they couldve been much worse!
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    Man this exam is going to be hard. They always throw in some really weird 1 and 2 markers here and there and I'm like how am I suppose to answer that...
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    Guys, can someone explain to me how you would normally do a rate of reaction question, where they give you a table of data showing concentrations then they tell you to find order of reaction for a particular reactant using the data and then find the rate equation.
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    Do we need to know the colour of EDTA Ligands? In both my revision guides the only ligand colour given to learn is [cu(nh3)4(h2o)2]^2+, but I did a past paper (can't remember which year) where it asked for the colour of an EDTA ligand?
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    (Original post by HoldtheDoor)
    Guys, can someone explain to me how you would normally do a rate of reaction question, where they give you a table of data showing concentrations then they tell you to find order of reaction for a particular reactant using the data and then find the rate equation.
    down a column the concentration of 1 reactant would usually change (I. E doubled) between 2 rows and UsUALLY the concentration of the other reactants would be kept the constant. This would in some way affect the rate (end column).

    If rate doubles when conc doubled, order w.r.t that chemical Is first. if it quadrupled order is 2nd. No change order=0

    Sometimes to get one of your orders they would have change 2 concentrations simultaneously, the trick is that you would be able to work out the order of one of the 2 reactants who's conc. Has been changed using the method mentioned prior and the other can be worked out with that new information if that makes sense :s

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    (Original post by Chloestar)
    Do we need to know the colour of EDTA Ligands? In both my revision guides the only ligand colour given to learn is [cu(nh3)4(h2o)2]^2+, but I did a past paper (can't remember which year) where it asked for the colour of an EDTA ligand?
    according to spec, only points on colours for f334 I think:
    recall that transition metals are d-block elements formingone or more stable ions which have incompletely filled dorbitals; recall the common oxidation states of iron andcopper and the colours of their aqueous ions; describe the colour changes in and write ionic equationsfor the reactions of: Fe2+(aq), Fe3+(aq) and Cu2+(aq) ionswith sodium hydroxide solution, and Cu2+(aq) ions withammonia solution;


    Also the iron(III) chloride test and potassium manganate titration colours

    copper II with ammonia is that colour you said
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    well i've done all the papers (the 2010 ones 3 times, the 2011s twice, june 2015 twice and the rest once)
    going to quickly rejig my memory on small parts of the course then go to bed, tomorrow morning maybe going through a paper or two again and then we'll see.

    despite the fact i've done the most papers for this exam than all of my other exams so far, i still feel so, so unprepared.
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    Not looking forward to this paper have no done very well in any of the practice papers is everyone scoring highly on previous years ?
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    Did anyone else find the Jun 2014 paper incredibly hard compared to the rest? I've been getting decent grades on other past papers but got a C in the jun 2014 :/
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    Saaaame
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    (Original post by lyricalvibe)
    down a column the concentration of 1 reactant would usually change (I. E doubled) between 2 rows and UsUALLY the concentration of the other reactants would be kept the constant. This would in some way affect the rate (end column).

    If rate doubles when conc doubled, order w.r.t that chemical Is first. if it quadrupled order is 2nd. No change order=0

    Sometimes to get one of your orders they would have change 2 concentrations simultaneously, the trick is that you would be able to work out the order of one of the 2 reactants who's conc. Has been changed using the method mentioned prior and the other can be worked out with that new information if that makes sense :s

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    Thanks! Can you elaborate on the last point you made.
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    (Original post by HoldtheDoor)
    Thanks! Can you elaborate on the last point you made.
    To figure out B we have to first realise the order wrt A is 1 as seen in row 2.

    Then we see in row 3 that doubling [A] and [B] means we have x4 the rate. We know doubling A accounts for a x2 to the rate tho so using this info,we realise B has added a further x2 ( to make a x4) while it's concentration was also doubled.

    Hence we deduce B is also order 1.

    C can be worked out in a similar way

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    Does anyone have any model answers?

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    Are we supposed to know why carboxylic acid is more acidic than alcohols? Like in terms of stability of the anions?
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    (Original post by kosvengali)
    Are we supposed to know why carboxylic acid is more acidic than alcohols? Like in terms of stability of the anions?
    Yh fam its in the spec but tbf there was a question on it in the jan 12 paper i think and the stability explanation was ignored. You were basically supposed to say how carboxylic acid is stronger than the rest because it can react with a strong base and form carbonate.
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    (Original post by lyricalvibe)
    To figure out B we have to first realise the order wrt A is 1 as seen in row 2.

    Then we see in row 3 that doubling [A] and [B] means we have x4 the rate. We know doubling A accounts for a x2 to the rate tho so using this info,we realise B has added a further x2 ( to make a x4) while it's concentration was also doubled.

    Hence we deduce B is also order 1.

    C can be worked out in a similar way

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    Thank you!
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Size:  499.4 KBWhy is the oxidation state of vanadium +4 for the first one?
 
 
 
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